For all real numbers x and y such that x \neq y, which of the following expressions is equ… — GRE General Test

For all real numbers

x

and

y

such that

x \neq y

, which of the following expressions is equivalent to

\frac{x^3 - x^2y - xy^2 + y^3}{x - y} - (x - y)^2

The equivalent expression is

2xy - 2y^2

Factoring the numerator by grouping gives

x^2(x - y) - y^2(x - y) = (x^2 - y^2)(x - y)

. Dividing by

(x - y)

yields

x^2 - y^2

. Expanding

(x - y)^2

gives

x^2 - 2xy + y^2

. Subtracting this expansion from

x^2 - y^2

results in

(x^2 - y^2) - (x^2 - 2xy + y^2) = 2xy - 2y^2

Factor the numerator of the rational fraction by grouping terms

x^3 - x^2y - xy^2 + y^3 = x^2(x - y) - y^2(x - y) = (x^2 - y^2)(x - y) = (x + y)(x - y)^2

Grouping terms allows factoring out common binomial components.

Simplify the fraction by dividing by

(x - y)

\frac{(x + y)(x - y)^2}{x - y} = (x + y)(x - y) = x^2 - y^2

Since

x \neq y

x - y \neq 0

, so common factors can be canceled.

Expand the subtracted term

(x - y)^2

(x - y)^2 = x^2 - 2xy + y^2

Applying the standard perfect square trinomial expansion formula.

Subtract the expanded expression from the simplified fraction result

(x^2 - y^2) - (x^2 - 2xy + y^2) = x^2 - y^2 - x^2 + 2xy - y^2 = 2xy - 2y^2

Distribute the negative sign across all terms inside the parentheses and combine like terms.

Simplifying algebraic expressions using factoring by grouping, difference of squares, and expanding binomial squares.

Question